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An invertiable matrix $A_{n\times n}$ is defined as there exists a matrix$B_{n\times n}$ such that $AB=BA=I_{n\times n}$.

Why isn't it defined as there exists a matrix$B_{n\times n}$ such that $AB=I_{n\times n}$? (a weaker condition)

Is it possible that $AB=I,\;BA\neq I$?

user3813057
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    As an aside, for infinite dimensional operators, it is indeed possible for $AB=I$ but $BA\neq I$. Take for example any sequence space, for example $\ell_2$ and the Right-shift and Left-shift operators, $\sigma_r$ and $\sigma_l$ respectively, where $\sigma_r((a_1,a_2,a_3,a_4,\dots))=(0,a_1,a_2,a_3,\dots)$ and $\sigma_l((a_1,a_2,a_3,a_4,\dots)=(a_2,a_3,a_4,a_5,\dots)$. One has $(\sigma_l\circ \sigma_r)((a_1,a_2,a_3,\dots))=(a_1,a_2,a_3,\dots)$ but $(\sigma_r\circ\sigma_l)((a_1,a_2,a_3,a)4,\dots))=(0,a_2,a_3,a_4,\dots)$ – JMoravitz Jan 09 '18 at 03:09

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